Question: Solve for $r$, $ \dfrac{4}{4r + 3} = -\dfrac{7}{16r + 12} - \dfrac{4r - 3}{16r + 12} $
Answer: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $4r + 3$ $16r + 12$ and $16r + 12$ The common denominator is $16r + 12$ To get $16r + 12$ in the denominator of the first term, multiply it by $\frac{4}{4}$ $ \dfrac{4}{4r + 3} \times \dfrac{4}{4} = \dfrac{16}{16r + 12} $ The denominator of the second term is already $16r + 12$ , so we don't need to change it. The denominator of the third term is already $16r + 12$ , so we don't need to change it. This give us: $ \dfrac{16}{16r + 12} = -\dfrac{7}{16r + 12} - \dfrac{4r - 3}{16r + 12} $ If we multiply both sides of the equation by $16r + 12$ , we get: $ 16 = -7 - 4r + 3$ $ 16 = -4r - 4$ $ 20 = -4r $ $ r = -5$